Arithmetic (UPSC) Part- IV

Total Questions: 50

31. A frog tries to come out of a dried well 4.5 m deep with slippery walls. Every time the frog jumps 30 cm, slides down 15 cm. What is the number of jumps required for the frog to come out of the well? [2020-II]

(a) 28
(b) 29
(c) 30
(d) 31

Correct Answer: (b) 29

Solution:Depth of the walls = 4.5 m = 450 cm.
Frog jump = 30 cm.
Frog slip = 15 cm.
Resultant height jump in one hop on = 30 – 15 = 15 cm.
Height jump in 28 hop on = 28 × 15 = 420 cm.
On 29th hop on, frog will jump,
420 + 30 = 450 cm or 4.5 m.

32. A bottle contains 20 litres of liquid A., 4 litres of liquid A is taken out of it and replaced by same quantity of liquid B. Again 4 litres of the mixture is taken out and replaced by same quantity of liquid B. What is the ratio of quantity of liquid A to that of liquid B in the final mixture? [2020-II]

(a) 4 : 1
(b) 5 : 1
(c) 16 : 9
(d) 17 : 8

Correct Answer: (c) 16 : 9

Solution:

	Liquid A	Liquid B
Initial	20	0
After first iteration	20 - 4 = 16	4

Ratio of Liquid A : Liquid B = 16 : 4 = 4 : 1.
When 4 litres of mixture taken out after first iteration.

Liquid A in 4 litre mix = 4 × (4/5) = 16/5

Liquid B in 4 litre mix = 4 × (1/5) = 4/5

Then, after second iteration,

Liquid A	Liquid B
16−16/5=64/5	4−4/5 + 4 = 36/5

33. The average score of a batsman after his 50th innings was 46.4. After 60th innings, his average score increases by 2.6. What was his average score in the last ten innings? [2020-II]

(a) 122
(b) 91
(c) 62
(d) 49

Correct Answer: (c) 62

Solution:Total runs scored by the batsman in 50 innings
= 50 × 46.4 = 2320.
Total runs scored by the batsman in 60 innings
= 60 × (46.4 + 2.6) = 2940.
Total runs scored in last 10 innings = 2940 – 2320 = 620.
Average score in last 10 innings = 620 / 10 = 62.

34. As a result of 25% hike in the price of rice per kg, a person is able to purchase 6 kg less rice for ₹1,200. What was the original price of rice per kg. [2020-II]

(a) ₹30
(b) ₹40
(c) ₹50
(d) ₹60

Correct Answer: (b) ₹40

Solution:Let original price of the rice is ₹ x/Kg.
Amount of rice purchased in ₹ 1200 = 1200 / x Kg.
When price increases by 25%, then price of rice is
₹ 1.25x/Kg.

and amount of rice purchased in ₹1200 = 1200 / 1.25x
ATQ,
1200 / x – 1200 / 1.25x = 6
1200(1.25 – 1) / 1.25x = 6 ⇒ x = 1200 × 0.25 / 1.25 × 6 = 40.
Hence, original price of rice is ₹ 40/Kg.

35. What is the greatest length x such that 3/2 m and 8/4 m are integral multiples of x? [2020-II]

(a) 1/2 m
(b) 1/3 m
(c) 1/4 m
(d) 3/4 m

Correct Answer: (d) 3/4 m

Solution:3(1/2) m = 7/2 m ⇒ 8(3/4) m = 35/4 m
For greatest length, we do H.C.F. of 7/2 and 35/4

= H.C.F. of 7 and 35 / L.C.M. of 2 and 4 = 7/4 = 1(3/4)

36. The recurring decimal representation 1.272727... is equivalent to [2020-II]

(a) 13/11
(b) 14/11
(c) 127/99
(d) 137/99

Correct Answer: (b) 14/11

Solution:1.272727..... = 127 – 1 / 99 = 126 / 99 = 14 / 11.

37. What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case? [2020-II]

(a) 1012
(b) 1022
(c) 1122
(d) 1222

Correct Answer: (b) 1022

Solution:L.C.M. of 3, 4, 5 and 6 = 60.
Least 4 digits number divided by 3, 4, 5 and 6 is
60 × 17 = 1020.
Remainder in each case = 2.
So, required number = 1020 + 2 = 1022.

38. In adult population of a city, 40% men and 30% women are married. What is the percentage of married adult population if no man marries more than one woman and no woman marries more than one man; and there are no widows and widowers? [2020-II]

(a) 33 × 1/7 %
(b) 34%
(c) 34 × 2/7 %
(d) 35%

Correct Answer: (c) 34 × 2/7 %

Solution:Let man adult population of the city is 300.
Then, married male adult population
= 300 × 40/100 = 120.

As, there are no widows and widowers and no man marries more than one woman and vice-versa.
So, number of married woman in the city = Number of married man in the city = 120.
ATQ,
Percent of married woman = 30%.

∴ Total woman population = 120 × 100 / 30 = 400.

Total population of the city = 300 + 400 = 700.
Total married population = 120 + 120 = 240.
Percent of married adult population
= 240/700 × 100 = 34(2/7)%

39. What is the remainder when 51 × 27 × 35 × 62 × 75 is divided by 100? [2020-II]

(a) 50
(b) 25
(c) 5
(d) 1

Correct Answer: (a) 50

Solution:51 × 27 × 35 × 62 × 75 ÷ 100
= 51 × 27 × 35 × 31 × 150 ÷ 100 = K × 150 ÷ 100
where K = 51 × 27 × 35 × 31
Now, when we divided 150 by 100,
remainder = 150 – 100 = 50.

40. A sum of ₹2,500 is distributed among X, Y and Z in the ratio 1/2 : 3/4 : 5/6 What is the difference between the maximum share and the minimum share? [2020-II]

(a) ₹300
(b) ₹350
(c) ₹400
(d) ₹450

Correct Answer: (c) ₹400

Solution:Ratio of shares of x : y : z = 1/2 : 3/4 : 5/6 = 6 : 9 : 10.

Shares of x = 2500 × 6/(6+9+10) = 600.

Shares of y = 2500 × 9/(6+9+10) = 900.

Shares of z = 2500 × 10/(6+9+10) = 1000.

Difference between maximum and minimum shares
= 1000 – 600 = ₹ 400.

« Previous 1 2 345 Next »